# Integrate the rational functions$\frac{1}{x^2-9}$

$\begin{array}{1 1}\large \frac{1}{6} \log|x-3|-\large \frac{1}{6} \log|x+3| + c \\ \frac{1}{6} \log|x-3|+ \frac{1}{6} \log|x+3| + c \\ \frac{1}{6} \log|x+3|-\large \frac{1}{3} \log|x-3| + c \\ \frac{1}{6} \log|x+3|- \frac{1}{6} \log|x+3| + c \end{array}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x-a)(x-b)},a\neq b$
• $(ii)\;$Form of the partial function$\frac{A}{x-a}+\frac{B}{x-b}$
• $(iii)\;\int\frac{dx}{x-a}=log(x-a)+c.$
Given $I=\int\frac{dx}{(x^2-9)}=\int\frac{dx}{(x-3)(x+3)}.$

This can be resolved as$\frac{A}{x-3}+\frac{B}{x+3}$

$\Rightarrow 1=A(x+3)+B(x-3).$

Equating the coefficients of x we get,

0=A+B-------(1)

Equating the constant terms

1=3A+3B------(2)

To solve for A and B,multiply the equ(1) by 3 and subtract from equ(2)

3A+3B=1------(1)
3A+3B=0-----(2)
_______________
$6A=1\Rightarrow A=\frac{1}{6}.$

substituting for A in equ(1) we get $B=\frac{-1}{6}.$

Hence $A=\frac{1}{6}$ and $B=\frac{-1}{6}.$

Now substituting for A and B in I we get,

Therefore $I=\frac{1}{6}\int\frac{dx}{x-3}-\frac{1}{6}\frac{dx}{x+3}.$

On integrating we get,

$\;\;\;=\frac{1}{6}log|x-3|-\frac{1}{6}log|x+3|$

log|a|-log|b|=log|a/b|,similarly

$I=\frac{1}{6}log\frac{|x-3|}{|x+3|}+c.$

answered Feb 5, 2013
edited Jul 21, 2013